Math Forum :: View topic – Closed+Compact=Closed?

I guess the result is typical. Though sequential approach is undoubtedly a very important tool in analysis, it may be inadequate when abstract topological spaces are under investigation. In particular, compactness and sequential compactness are not equivalent in general. So I would like to reproduce the proof here as it’s “sequence–free”. hope the community will be interested.

Firstly, by the continuity of the operation , for all neighbourhood of the identity , there is an neighbourhood of of , such that and

Suppose now that is compact and is an neighbourhood containing . Fixed . Again, by the continuity of , there is a neighbourhood of such that . In view of the first observation, we can find also neighbourhood of , such that As a result is a neighbourhood of and By the compactness of , there exists such that Set and it could be checked that

Now let but Then As is compact and is open, by the above assertion with and , we can find a neighbourhood of such that That is, , which is equivalent to ! Contradicting that

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